Answer to Question #210380 in Mechanical Engineering for arif

Question

Draw the profile of the cam when (a) the line of stroke of the valve rod passes through the axis of the cam shaft, and (b) the line of the stroke is offset 20 mm from the axis of the cam shaft. The cam shaft has maximum velocity and acceleration at 100 rpm and during valve raised and lowered have 0.30 m/s; 0.725 m/s and 6.13 m/s2; 24.67 m/s2 respectively

Answer 1

The profile of the cam is drawn as discussed in the following steps :

1. Draw a base circle with center O and radius equal to the minimum radius of the cam ( i.e., 25 mm ).

2. Draw a prime circle with center O and radius,

$OA= Min. radius of cam + 2 1 ​ Dia. of roller =25+ 2 1 ​ ×20=35mm$

3. Draw angle AOS = 120^∘

 to represent raising or out stroke of the valve, angle SOT = 30^∘

 to represent dwell and angle TOP = 60^∘

 to represent lowering or return stroke of the valve.

4. Divide the angular displacements of the cam during raising and lowering of the valve (i.e., angle AOS and TOP ) into the same number of equal even parts as in the displacement diagram.

5. Join the points 1, 2, 3, etc., with the center O and produce the lines beyond the prime circle as shown in Fig. 20.17.

6. Set off 1B, 2C, 3D, etc., equal to the displacements from the displacement diagram.

7. Join the points A, B, C … N, P, A. The curve drawn through these points is known as the pitch curve.

8. From the points A, B, C … N, P, draw circles of radius equal to the radius of the roller.

9. Join the bottoms of the circles with a smooth curve. This is the required profile of the cam.

The profile of the cam when the line of stroke is offset from the axis of the camshaft, as shown in Fig. 20.18, maybe drawn as discussed in the following steps :

1. Draw a base circle with center O and radius equal to 25 mm.

2. Draw a prime circle with center O and radius OA = 35 mm.

3. Draw an offset circle with center O and radius equal to 15 mm.

4. Join OA. From OA draw the angular displacements of cam, i.e., draw angle AOS = 120^∘

, angle SOT = 30^∘ and angle TOP =60^∘

5. Divide the angular displacements of the cam during raising and lowering the valve into the same number of equal even parts (i.e., six parts ) as in the displacement diagram.

6. From points 1, 2, 3 …. etc. and 0′,1′, 3′, …etc. On the prime circle, draw tangents to the offset circle.

7. Set off 1B, 2C, 3D… etc., equal to displacements as measured from displacement diagram.

8. By joining the points A, B, C … M, N, P, we get a pitch curve with a smooth curve.

9. Now, A, B, C…etc. as the center, draw circles with radius equal to the radius of roller.

10. Join the bottoms of the circles with a smooth curve. This is the required profile of the cam.

Maximum acceleration of the valve rod

We know that angular velocity of the camshaft,

$\omega=\frac{2 \pi N}{60}=\frac{2 \pi \times 100}{60}=10.47 rad / s$

We also know that maximum velocity of the valve rod to raise valve,

$v_{ O }=\frac{\pi \omega . S}{2 \theta_{ O }}=\frac{\pi \times 10.47 \times 0.05}{2 \times 2.1}=0.39 m / s$

and maximum velocity of the valve rod to lower the valve,

$v_{ R }=\frac{\pi \omega . S}{2 \theta_{ R }}=\frac{\pi \times 10.47 \times 0.05}{2 \times 1.047}=0.785 m / s$

The velocity diagram for one complete revolution of the cam . We know that the maximum acceleration of the valve rod to raise the valve.

$a_{ O }=\frac{\pi^{2} \omega^{2} \cdot S}{2\left(\theta_{0}\right)^{2}}=\frac{\pi^{2}(10.47)^{2} 0.05}{2(2.1)^{2}}=6.13 m / s ^{2}$

and maximum acceleration of the valve rod to lower the valve,

$a_{ R }=\frac{\pi^{2} \omega^{2} \cdot S}{2\left(\theta_{ R }\right)^{2}}=\frac{\pi^{2}(10.47)^{2} 0.05}{2(1.047)^{2}}=24.67 m / s ^{2}$

The acceleration diagram for one complete revolution of the cam